If the work done on the system or by the system is zero, which one of the following statements for a gas kept at a certain temperature is correct?
- (a)Change in internal energy of the system is equal to flow of heat in or out of the system.
- (b)Change in internal energy of the system is less than heat transferred.
- (c)Change in internal energy of the system is more than the heat flow.
- (d)Cannot be determined.
Correct - A, the change in internal energy equals the heat flowing in or out. The first law of thermodynamics is Delta U = Q - W, where Q is the heat added to the system and W is the work done BY the system. If no work is done (W = 0), the equation reduces to Delta U = Q: every joule of heat entering or leaving the gas goes entirely into changing its internal energy. So the change in internal energy is exactly equal to the heat flow.
- (b)Change in internal energy of the system is less than heat transferred. — For the internal-energy change to be less than the heat transferred, the gas would have to do positive work (W greater than 0); but the question fixes W = 0, so no heat is diverted into work.
- (c)Change in internal energy of the system is more than the heat flow. — For the change to exceed the heat flow, work would have to be done ON the gas to add extra internal energy, again contradicting W = 0.
- (d)Cannot be determined. — With W = 0 the first law gives a definite result, Delta U = Q, so the change is fully determined, not indeterminate.
The first law of thermodynamics is a statement of energy conservation for a gas: the change in internal energy equals the heat added minus the work done by the gas (Delta U = Q - W). Internal energy is a state property, depending for an ideal gas only on temperature. When the process does zero work, the heat exchanged and the internal-energy change are numerically identical.
Identify what is zero. Setting the work done on or by the system to zero removes the W term and leaves Delta U = Q. This is the constant-volume (isochoric) case, where all the heat changes the internal energy.
- The first law states Delta U = Q - W, with Q the heat added and W the work done by the system.
- If W = 0, then Delta U = Q.
- Zero work typically corresponds to a constant-volume (isochoric) process.
- For an ideal gas, internal energy depends only on temperature.
With W = 0 the first law reduces to Delta U = Q - option (a).
- Getting the sign of W wrong - here W is the work done BY the gas.
- Forgetting that a constant-volume process does zero work, so all heat becomes internal energy.
Asked by applying the first law to a process where either the work or the heat is zero, or by identifying the constant-volume case.
No directly related past PYQ was found.
- practice — not a real PYQ
In a constant-volume process, the heat supplied to a gas goes entirely into
- (a)work done by the gas
- (b)change in internal energy
- (c)change in pressure only
- (d)latent heat
Answer(b) change in internal energy - no work is done at constant volume.
- practice — not a real PYQ
The first law of thermodynamics is a statement of the conservation of
- (a)charge
- (b)momentum
- (c)energy
- (d)mass
Answer(c) energy - it balances heat, work and internal energy.